ПОМОГИТЕ ! СРОЧНО !......

Решите задачу:

$1)f(x)=2-\frac{2}{\sqrt{x} }=2-2x^{-\frac{1}{2} }\\\\f'(x)=2'-2(x^{-\frac{1}{2} })'=-2*(-\frac{1}{2}x^{-\frac{3}{2} })=\frac{1}{\sqrt{x^{3} } } =\frac{1}{x\sqrt{x} } \\\\2)f(x)=3-\frac{1}{Sin^{2}x } =3-Sin^{-2} x\\\\f'(x)=3'-(Sin^{-2}x)'=2Sin^{-3}x*(Sinx)'=2Sin^{-3}x*Cosx=\frac{2Cosx}{Sin^{3} x}$

$3)f(x)=\frac{x^{3}-1 }{x^{2} } \\\\f'(x)=\frac{(x^{3}-1)'*x^{2}-(x^{3}-1)*(x^{2})'}{(x^{2})^{2}}=\frac{3x^{2}*x^{2}-(x^{3}-1)*2x }{x^{4} }=\frac{3x^{4}-2x^{4}+2x }{x^{4} }=\frac{x^{4}+2x }{x^{4} }=\frac{x^{3} +2}{x^{3} }$